Difference between revisions of "2005 AMC 10A Problems/Problem 6"
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Therefore, the average of all <math>50</math> numbers is <math>\frac{1200}{50}=24 \Longrightarrow \mathrm{(B)}</math> | Therefore, the average of all <math>50</math> numbers is <math>\frac{1200}{50}=24 \Longrightarrow \mathrm{(B)}</math> | ||
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==See also== | ==See also== | ||
Revision as of 15:29, 8 December 2020
Problem
The average (mean) of 
numbers is 
, and the average of other numbers is . What is the average of all 
numbers?
Solution
Since the average of the first numbers is , their sum is 
.
Since the average of other numbers is , their sum is 
.
So the sum of all numbers is 
Therefore, the average of all numbers is 
See also
| 2005 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 5 |
Followed by Problem 7 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
| All AMC 10 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
